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No-one
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If $\mathcal{H}^{n-1}(K)=0$ in $\mathbb{R}^n$ then $\mathcal{H}^n(K\times \mathbb{R})=0$ in $\mathbb{R}^{n+1}$

I am reading a paper where the authors seem to claim that if $K\subset\mathbb{R}^n$ is a compact set with $\mathcal{H}^{n-1}(K)=0$ in $\mathbb{R}^n$ ($\mathcal{H}^{n-1}$ is the $(n-1)$-dimensional Hausdorff measure), then $\mathcal{H}^n(K\times \mathbb{R})=0$ in $\mathbb{R}^{n+1}$.

I want to know how to prove this.

For example, I would be able to prove it if I knew how to prove the following: If $K\subset \mathbb{R}^n$ is a compact set such that $\mathcal{H}^s(K)=0$ ($\mathcal{H}^s$ is the $s$-dimensional Hausdorff measure) for some $s<n$, then for all $\epsilon>0$ $K$ can be covered by finitely many $B_r(x_i)$, $i=1,...,N$ (same radius) such that $N r^s<\epsilon$.

Any hints are appreciated.

No-one
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